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Quantum Mechanics - Angular Momentum probability

  • Thread starter Tangent87
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  • #1
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See the attachment for the question.

I'm stuck on the bit highlighted in yellow (have done the rest of the question).

I'm not sure if it's really trivial or more involved. Do I have to use the fact that U(pi/2)|j m> are eigenvectors of J1 with eigenvalue m somehow? Why is the answer not just:
|<j m'| J1 |j m>|2?

Thanks
 

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  • #2
vela
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You don't generally apply the operator when you're making a measurement. Instead, what you do is expand the state of the system in terms of the eigenstates of the observable

[tex]\vert j m \rangle = \sum_{m' = -j}^j c_{m'} \vert j m' \rangle_{J_1}[/tex]

(where [itex]\vert j m' \rangle_{J_1}[/itex] are the eigenstates of J1) and then pick off the coefficients (taking advantage of the orthonormality of the eigenstates) and square them to calculate the probabilities.
 
  • #3
148
0
You don't generally apply the operator when you're making a measurement. Instead, what you do is expand the state of the system in terms of the eigenstates of the observable

[tex]\vert j m \rangle = \sum_{m' = -j}^j c_{m'} \vert j m' \rangle_{J_1}[/tex]

(where [itex]\vert j m' \rangle_{J_1}[/itex] are the eigenstates of J1) and then pick off the coefficients (taking advantage of the orthonormality of the eigenstates) and square them to calculate the probabilities.

Ah yes of course! I've done it now, cheers.
 
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